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alisha [4.7K]
3 years ago
11

The function f(x)=603(1.3)^x represents the number of students enrolled at a university x years after it was founded. Each year,

the number is students is______ the number the year before.
Mathematics
2 answers:
fredd [130]3 years ago
6 0

Answer:

1.3 times

Step-by-step explanation:

The function f(x)=603(1.3)^{x} represents the number of students enrolled at a university, x years after it was founded.

So the sequence will be formed to represent the number of students will be

f(1) = 603(1.3)

f(2) = 603(1.3)²

f(3) = 603(1.3)³

and so on.

Now the common ratio between second and first term will be

= \frac{(603)(1.3)^{2}}{603(1.3)}=1.3

Therefore, second term of the sequence will be 1.3 times of the first term.

Answer will be - "Each year, the number of students will be 1.3 times the number the year before".

Ede4ka [16]3 years ago
5 0

Answer: 0.3 times

Step-by-step explanation:

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8 0
3 years ago
There are 450 total students at Snowden School. On Tuesdays, 12% of the students stay after school for Comedy Club. How many stu
Anestetic [448]

Answer:

54 students

Step-by-step explanation:

First, change the percent to a decimal.

12% = 0.12

Now, multiply the decimal by the total amount of students to find the amount of students that stay after school.

450 × 0.12 = 54

54 students stay after school for Comedy Club.

6 0
3 years ago
Read 2 more answers
Tell whether -2 is a solution of the inequality n-5<8
ruslelena [56]

Answer:

-2 is a solution

Step-by-step explanation:

-2-5<8

-7<8

6 0
3 years ago
A public swimming pool that holds 45,000 gallons of water is going to be drained for maintenance at a rate of 100 gallons per mi
crimeas [40]

Answer:

450 minutes

Step-by-step explanation:

Given

w = 45000 - 100t

Required

Determine how long it'll take to empty the pool

Being empty means there's no water in the pool.

So, we have to set w to 0 to get the value of t

i.e w = 0

Substitute 0 for w in w = 45000 - 100t

0 = 45000 - 100t

Collect Like Terms

100t = 45000

Divide through by 100.

t = 450

Hence, it'll take 450 minutes to empty the pool

8 0
3 years ago
A sample of 11001100 computer chips revealed that 62b% of the chips fail in the first 10001000 hours of their use. The company's
STALIN [3.7K]

Answer:

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 90% confidence interval) ---- reject Null hypothesis

Z score < Z(at 90% confidence interval) ------ accept Null hypothesis

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

z score = 1.35

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

Question; A sample of 1100 computer chips revealed that 62% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 60% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.10 level.

Step-by-step explanation:

Given;

n=1100 represent the random sample taken

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 1100

po = Null hypothesized value = 0.60

p^ = Observed proportion = 0.62

Substituting the values we have

z = (0.62-0.60)/√{0.60(1-0.60)/1100}

z = 1.354

z = 1.35

To determine the p value (test statistic) at 0.01 significance level, using a two tailed hypothesis.

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

3 0
3 years ago
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